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I've read that two polynomials $r_1, r_2 \in R[X]$ on the form $r = a_nX^n + ... + a_1X + a_0$ are equal if and only if the cofficients of $r_1, r_2$: $a_i, b_i$ are equal for all $i, 0 \leq i \leq n$. Here $R[X]$ denote the polynomial ring on $R$, where $R$ is a ring.

Is this purely a definition or can I proof this ? Can two polynomials with different cofficients achieve the same values for all $x \in R$ ?

I can write $a_nX^n + ... a_1X + a_0 = b_nX^n + ... b_1X + b_0$ set $X = 0$ and deduce $a_0 = b_0$. Then $a_nX^n + ... a_1X = b_nX^n + ... b_1X$ implies $X(a_nX^{n-1} + ... a_1) = X(b_nX^{n-1} + ... b_1)$ but now I can't conclude $a_1 = b_1$ ?

Thanks

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2 Answers 2

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Purely by definition, as @Magdiragdag pointed out.

But beware: the subtlety here is that polynomials and polynomials functions are not necessarily the same thing!

Polynomials can be turned into functions if you "plug in" values for X. But if the values for X lies in a finite field then it can happen that different polynomials give rise to the same polynomial function!

For example, take the field $Z_2 = \{0,1\}$ of residues mod 2, then $X^2$ and $X$ are different polynomials in $Z_2[X]$ (they have different coefficients) but the corresponding polynomial functions are the same, since $X^2 = X$ for $X = 0$ and $1$. More generally, the different polynomials $X^p$ and $X$ give rise to the same polynomial function for $X$ lying in the field $Z_p$ of residues mod p (p a prime), as an immediate consequence of Fermat's Little Theorem.

To go on with your argument and conclude that $a_1 = b_1$ in the polynomials seen as polynomial functions over R you must be able to choose a nonzero X in the field R which is not a root of neither $a_n X^{n−1}+\cdots+a_1$, nor $b_n X^{n−1}+\cdots+b_1$, which you can do safely if your ring R is an infinite field, such as the rationals, the reals or the complex numbers.

So, following your proof, for infinite fields R we get that polynomials in R[X] and polynomials functions with domain R are "the same thing". But this is not true for an arbitrary ring R, as the examples above point out.

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  • $\begingroup$ Thanks a lot for your comment Lucas. $\endgroup$
    – Shuzheng
    Oct 16, 2013 at 19:15
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Yes, this is purely by definition. If you were to formally define/construct polynomials, you'd define $R[X]$ as the set of functions $r \colon {\mathbb N} \to R$ that have only finitely many non-zero elements (equipped with the suitable structure of a ring). The values $r(i)$ are the ''coefficients'' and two such functions are equal if all their ''coefficients'' are equal.

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    $\begingroup$ How does the formal definition of a polynomial relate to 'real world' polynomial $f(X) = a_nX^n + ... + a_1X + a_0$ ? I mean, the difference seems to be huge ? Why are we not writing polynomials in the formal way ? And why define something formal if concrete instances doesn't reflect the formalism ? $\endgroup$
    – Shuzheng
    Oct 16, 2013 at 17:48
  • $\begingroup$ @NicolasLykkeIversen, that's because polynomials and polynomial functions are not necessarily the same thing. See if my answer below helps. $\endgroup$
    – Lucas Seco
    Oct 16, 2013 at 18:09
  • $\begingroup$ @NicolasLykkeIversen the relation is that the function $r \colon {\mathbb N} \to R$ corresponds to the 'real world' polynomial $r(0) + r(1)X + r(2)X^2 + \dots$ (which has finite degree, because $r$ has finite carrier). So $r$ is just a way of writing down the coefficients of a polynomial. $\endgroup$ Oct 16, 2013 at 18:47
  • $\begingroup$ Thanks a lot. So we define polynomials using $r: \mathbb N \rightarrow R$ - This function is like the signature/identity of the polynomial ? And when we substitute $X$ for some $r\in R$ we are turning our polynomial into a polynomial function ? A polynomial is just an algebraic object itself, and can be turned into a polynomial function upon instantiate? $\endgroup$
    – Shuzheng
    Oct 16, 2013 at 19:11
  • $\begingroup$ Exactly @NicolasLykkeIversen! $\endgroup$
    – Lucas Seco
    Oct 16, 2013 at 19:52

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